Prof that it is not possible to get a standing wave.

[center o bass]

Homework Statement

As in the Title starting from two planewaves with the same amplitude, but different frequency.

Homework Equations

Starting from
$$Ae^{i(kx +\omega_1 t)} + Ae^{i(kx +\omega_2 )}$$

The Attempt at a Solution

I got as far as

$$Ae^{i(kx +\omega_1 t)} + Ae^{i(kx +\omega_2 )}$$
$$= 2Ae^{ikx} e^{i \frac{\omega_2 + \omega_1}{2} t} \cos \frac{\omega_2 - \omega_1}2 t$$

taking the real part

$$= 2A \left( \cos kx \cos{ \frac{\omega_2 + \omega_1}{2} }t - \sin kx \sin { \frac{\omega_2 + \omega_1}{2} } \right) \cos{ \frac{\omega_2 - \omega_1}{2}}$$

Now how do i argue to conclude that this is definitley not a standing wave? I know that position and time has to be decoupled, but how can I conclude that they are not?

Can this expression be simplified further?

is it enough to say that this is a product of two factors one which depends only on time and one which does depend on both space and time and therefore can not be decoupled?

 

[kuruman]

I would start with an expression that reduces to a standing wave when the frequencies are the same. Does your starting expression do that?

 

[center o bass]

Yes it does when $$\omega_1 = - \omega_2$$

 

[kuruman]

By your logic, if ω₁ = 100 Hz, then ω₂ = -100 Hz. Is 100 Hz equal to -100 Hz? Do it right and put the minus sign where it belongs. Wave frequencies are always positive.